what is the limit as x approaches negative infinity of (4x/x-6)+(5x/x+6) ?

Question

sorry it's not negative infinity. is it still 9 tho?

Answer 1

I will give you an example and you solve your question.

When the degree of the numerator is the same as that of the denominator, the limit is the ratio of the leading co-efficient s(ie, co-efficient of highest power of x)

Example: lim x---> -inf (3x+ 2)/ (-2x+1)
= -3/2.
Now find the limit of each rational function in the sum and add to get the result.

At + infinity or minus infinity, it is ratio of leading co-efficients. At other points in the domain of the function, the limit is found by substituting.
Example:
lim x----->5 [(x-1)/(x+2)] = 4/7

If we are finding the limit at x=-2, the number -2 makes the denominator 0, without making the numerator 0. Hence the function will tend to plus or minus infinity, depending on the direction from which we are approaching -2.

From the right, the numerator will be negative and denominator positive, hence it will go to minus infinity. From the left, it will go to infinity.

Answer 2

I guess you mean 4x/(x-6)+5x/(x+6) . If you divide by x, this can be rewritten as 4/(1-6/x) + 5/(1 + 6/x). When x-> -oo, 1-6/x -> 1 and 1+ 6/x -> 1, because 1/x -> 0.

So, you limit is 4/1 + 5/1 = 9

If it is negative infinity, it's still 9

Answer 3

lim (4x/x-6)+(5x/x+6)
= lim ((4x(x+6))/((x+6)(x-6)) + (5x(x-6))/((x+6)(x-6)))
= lim ((4x^2+24x)/((x+6)(x-6)) + (5x^2-30x))/((x+6)(x-6)))
= lim ((9x^2-6x)/(x^2-36)
= lim ((9-6/x)/(1-36/x^2))

the limit as x approachs infinity or negative infinity of n/x or n/x^2 is 0 ... think of it as a small number over a REALLY big number

= lim (9-0)/(1-0) = lim 9 = 9

Here is what I did, I rewrote it to 1 term. That is something you should do immediatly. Then I divided by the greatest power of x(in this case it was x squared). I figure lim 1/x = 0 so i replaced anthing over x with 0 and solved! The answer is 9 :) Good Luck!

Answer 4

Do you by any chance mean:
[ 4x / (x - 6) ] + [ 5x / (x + 6) ] ?
If so, divide the numerator and denominator of each fraction by x, obtaining:
[ 4 / (1 - 6/x) } + [ 5 / (1 + 6/x) ].
As x approaches -infty, the two expressions involving 6/x tend to zero, leaving:
4 / 1 + 5 / 1
= 9.

Answer 5

Since there will be inf/inf when you plug -inf directly on it. This is indeterminate, using Le Hospitals' Theorem take the derivative on both numerator and denominator for each.
Then plug back the -inf limit.

Therefore, derivative of 4x is 4 and derivative of x-6 is 1. And so on with the other. Then the limits will become,

4/1 + 5/1 which is 9.

Answer 6

Ans: 9

As x -> - inf,

The denominator will look like x, since -6 and +6 is insignificant in comparison to negative infinity.

So, the expression will become:

(4x/x) + (5x/x) = 4+5 = 9

Answer 7

first element is to remember general shrink regulations do not persist with to purposes coping with infinity. So what to do then? you desire examine out the function and discover the optimal ability of x. (x^2) Then divide the finished function by potential of the optimal ability So we could get: (a million/x) -------------------- (sixteen/x^2) + a million ok now we are able to cancel out (a million/x) and (sixteen/x^2) with the aid of fact if we plug in an somewhat super destructive extensive variety (ex. -a million billion) for x for (a million/x) we could get -.000000001 which rounds to 0 and we could get an identical answer for (sixteen/x^2). so as that leaves us with: 0/0+a million = 0 which potential as x gets infinitely greater advantageous in a destructive course f(x) gets infinitely closer and closer to 0. the version between this and the previous strategies is the quantity of artwork you are able to teach. some instructors will require you to teach not basically the best judgment however the algebra to boot. examine at the same time with your instructor and notice with technique his or she will settle for. i did not placed the algebra on one among my try to the instructor docked me 2 or 3 factors for the concern.